(1) $P=\left(\begin{array}{ccc}\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ -\frac{1}{\sqrt{3}} & 0 & \frac{2}{\sqrt{6}}\end{array}\right)$; (2) $C=\left(\begin{array}{ccc}\frac{5}{3} & -1 & -1 \\ -1 & \frac{5}{3} & \frac{1}{3} \\ -1 & \frac{1}{3} & \frac{5}{3}\end{array}\right)$.
\section*
(1) 由 $|\lambda E-A|=\left|\begin{array}{ccc}\lambda-a & -1 & 1 \\ -1 & \lambda-a & 1 \\ 1 & 1 & \lambda-a\end{array}\right|=(\lambda-a+1)^{2}(\lambda-a-2)=0$
得 $\lambda_{1}=a+2, \lambda_{2}=\lambda_{3}=a-1$
当 $\lambda_{1}=a+2$ 时
$((a+2) E-A)=\left(\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & 1 \\ 1 & 1 & 2\end{array}\right) \stackrel{r}{\rightarrow}\left(\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0\end{array}\right)$ 的特征向量为 $\alpha_{1}=\left(\begin{array}{c}1 \\ 1 \\ -1\end{array}\right)$,
当 $\lambda_{2}=\lambda_{3}=a-1$ 所
$((a-1) E-A)=\left(\begin{array}{ccc}-1 & -1 & 1 \\ -1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right) \stackrel{\underset{\sim}{1}}{r}\left(\begin{array}{ccc}1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right)$ 的特征向量为 $\alpha_{2}=\left(\begin{array}{c}-1 \\ 1 \\ 0\end{array}\right), \alpha_{3}=\left(\begin{array}{c}-1 \\ 1 \\ 2\end{array}\right)$,
令 $P=\left(\frac{\alpha_{1}}{\left|\alpha_{1}\right|}, \frac{\alpha_{2}}{\left|\alpha_{2}\right|}, \frac{\alpha_{3}}{\left|\alpha_{3}\right|}\right)=\left(\begin{array}{ccc}\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ -\frac{1}{\sqrt{3}} & 0 & \frac{2}{\sqrt{6}}\end{array}\right)$, 则 $P^{T} A P=\Lambda=\left(\begin{array}{lll}a+2 & & \\ & a-1 & \\ & & a-1\end{array}\right)$,
(2) $\left.P^{T} C^{2} P=P^{T}(a+3) E-A\right) P=\left((a+3) E-\Lambda=\left(\begin{array}{lll}1 & & \\ & 4 & \\ & & 4\end{array}\right)\right.$
$\Rightarrow P^{T} C P P^{T} C P=\left(\begin{array}{lll}1 & & \\ & 4 & \\ & & 4\end{array}\right) \Rightarrow P^{T} C P=\left(\begin{array}{lll}1 & & \\ & 2 & \\ & & 2\end{array}\right)$,
故 $C=P\left(\begin{array}{lll}1 & & \\ & 2 & \\ & & 2\end{array}\right) P^{T}=\left(\begin{array}{ccc}\frac{5}{3} & -1 & -1 \\ -1 & \frac{5}{3} & \frac{1}{3} \\ -1 & \frac{1}{3} & \frac{5}{3}\end{array}\right)$.